Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 1}{x - 6} = \dfrac{-5x + 5}{x - 6}$
Answer: Multiply both sides by $x - 6$ $ \dfrac{x^2 - 1}{x - 6} (x - 6) = \dfrac{-5x + 5}{x - 6} (x - 6)$ $ x^2 - 1 = -5x + 5$ Subtract $-5x + 5$ from both sides: $ x^2 - 1 - (-5x + 5) = -5x + 5 - (-5x + 5)$ $ x^2 - 1 + 5x - 5 = 0$ $ x^2 - 6 + 5x = 0$ Factor the expression: $ (x - 1)(x + 6) = 0$ Therefore $x = 1$ or $x = -6$ The original expression is defined at $x = 1$ and $x = -6$, so there are no extraneous solutions.